Scala，具有值列表的 Dataframe 列，我想将每个值创建为新列并想命名它

【问题标题】：Scala , Dataframe column with a list of values, I want to create each value as new column & want to name itScala，具有值列表的 Dataframe 列，我想将每个值创建为新列并想命名它
【发布时间】：2018-11-15 20:55:14
【问题描述】：

我在下面给出了一个数据框列。

House_No = INT
family_details = ["name" , age , "surname" , weight]
Ownership = Boolean

我想为数据框创建包含姓名、年龄、姓氏和体重的新列。

House_No
family_details
Ownership
name
age
surname
weight

【问题讨论】：

标签： scala functional-programming apache-spark-sql user-defined-functions

【解决方案1】：

以下解决方案将为您提供帮助：

     val data =  Array((2,Array("abc","23","xyz","70"),true),(3,Array("lmn","45","pqr","50"),false))

     val rdd = sc.parallelize(data)

     val df = rdd.toDF("house_no","family_details","ownership")

val res = df.select("house_no","ownership","family_details").withColumn("name", split($"family_details" (0), ",")(0)).withColumn("age", split($"family_details"(1), ",")(0)).withColumn("surmname", split($"family_details"(2), ",")(0)).withColumn("Weight", split($"family_details"(3), ",")(0)).drop("family_details")

【讨论】：